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SVEN [57.7K]
3 years ago
10

The masses of the two moons are determined to be 2M2M for Moon AA and MM for Moon BB . It is observed that the distance between

Moon BB and the planet is two times that of the distance between Moon AA and the planet. How does force exerted from the planet on Moon AA compare to the force exerted from the planet on Moon BB
Physics
2 answers:
seraphim [82]3 years ago
7 0

Answer:

 F_A = 8 F_B

Explanation:

The force exerted by the planet on each moon is given by the law of universal gravitation

        F = G \frac{m M}{r^{2} }

where M is the mass of the planet, m the mass of the moon and r the distance between its centers

let's apply this equation to our case

Moon A

the distance between the planet and the moon A is r and the mass of the moon is 2m

        F_A = G \frac{2m M}{r^{2} }

Moon B

        F_B = G \frac{m M}{(2r)^{2} }

         F_B = G \frac{m M}{4 r^{2} }

the relationship between these forces is

         F_B / F_A = \frac{1}{2 \ 4 } = 1/8

         F_A = 8 F_B

Kobotan [32]3 years ago
7 0

Answer:

F_A = 8 F_B

Explanation:

The force exerted by the planet on each moon is given by the law of universal gravitation

       F =

where M is the mass of the planet, m the mass of the moon and r the distance between its centers

let's apply this equation to our case

Moon A

the distance between the planet and the moon A is r and the mass of the moon is 2m

       F_A = G \frac{2m M}{r^{2} }

Moon B

       F_B = G \frac{m M}{(2r)^{2} }

        F_B = G \frac{m M}{4 r^{2} }

the relationship between these forces is

        F_B / F_A =  = 1/8

        F_A = 8 F_B

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Hunter-Best [27]

The car will take 300 m before it stops due to applying break.

<h3>What's the relation between initial velocity, final velocity, acceleration and distance?</h3>
  • As per Newton's equation of motion, V² - U² = 2aS
  • V= final velocity velocity of the object, U = initial velocity velocity of the object, a= acceleration, S = distance covered by the object
  • Here, U = 60 ft/sec, V = 0 m/s, a= -6 ft/sec²
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=> -3600 = -12S

=> S = 3600/12 = 300 m

Thus, we can conclude that the distance covered by the car is 300 m before it stopped.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: A car is being driven at a rate of 60 ft/sec when the brakes are applied. The car decelerates at a constant rate of 6 ft/sec². How long will it take before the car stops?

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2 years ago
The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.50×10−9 m2 , a plate separ
Marizza181 [45]

The energy stored in the membrane is 6.44\cdot 10^{-14} J

Explanation:

The capacitance of a parallel-plate capacitor is given by

C=\frac{k\epsilon_0 A}{d}

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k is the dielectric constant of the material

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the membrane in this problem, we have

k = 4.6

A=4.50\cdot 10^{-9} m^2

d=8.1\cdot 10^{-9} m

Substituting, we find its capacitance:

C=\frac{(4.6)(8.85\cdot 10^{-12})(4.50\cdot 10^{-9})}{8.1\cdot 10^{-9}}=2.26\cdot 10^{-11} F

Now we can find the energy stored: for a capacitor, it is given by

U=\frac{1}{2}CV^2

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C=2.26\cdot 10^{-11} F is the capacitance

V=7.55\cdot 10^{-2} V is the potential difference

Substituting,

U=\frac{1}{2}(2.26\cdot 10^{-11} F)(7.55\cdot 10^{-2})^2=6.44\cdot 10^{-14} J

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6 0
3 years ago
if the mass is 10 gram and the volume is 7 cubic centimetre,find the density in kilogram per cubic metre
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Answer:

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How to derive the fourth equation of motion?
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Answer:

To derive the fourth equation of motion, first we have to consider the equation for acceleration and then to rearrange it. or v2 = u2 + 2as and this equation of motion can be used to find the final velocity or the distance travelled if the other values are given.

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Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

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0 = 625 + 100A....then moved the other value to one

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(ii) Firstly convert 54km/hr to m/s

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then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

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Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

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3 years ago
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